NDA Maths · Probability

Classical Probability & Counting

Probability as a counting ratio — favourable outcomes over total — applied to cards, dice, coins, arrangements, and number selections.

All Probability notes NDA Maths strategy

Why this matters

Start here. Every other probability subtopic — event algebra, independence, conditional probability, Bayes — is built on the classical idea below: count the favourable outcomes, count the total, divide. This is the single largest slice of the NDA Probability bank (85 questions across dice, coins, balls, arrangements, and number-selection problems), and the easy marks live here — roughly a third of the bank is EASY. Master the counting (combinations, sample-space size, the complement trick) and the rest of the chapter becomes bookkeeping.

Concept 1 of 8

What is probability? (Random experiments, sample space, events)

Intuition

A random experiment is any process whose individual result you cannot predict in advance, even though you know every result that COULD occur — tossing a coin, rolling a die, drawing a card. Each possible result is an outcome; the set of all outcomes is the sample space; any collection of outcomes you care about is an event. Probability measures how much of the sample space an event covers.

Definition

Four pieces of vocabulary the whole chapter rests on:

A random experiment has a known set of possible results but an unpredictable individual result.
The sample space $S$ is the set of all outcomes — for one die $S=\{1,2,3,4,5,6\}$ .
An event $E$ is any subset $E \subseteq S$ ; it occurs when the actual outcome lies in $E$ .
Outcomes are equally likely when symmetry gives no reason to prefer one over another (a fair die, a fair coin) — the assumption the classical definition rests on.

Diagram · event = subset of the sample space

P(E) = favourable / total = 3 / 6 = 1/2

The sample space S is all six equally likely outcomes; the event E is the subset {4, 5, 6}. For equally likely outcomes, P(E) is simply the number of favourable outcomes over the total.

Worked example

Two coins are tossed together. Write the sample space

S

, then write the event

A

= "at least one head" and count its outcomes.

List every ordered outcome of the two coins: $S = \{HH,\ HT,\ TH,\ TT\}$ , so $n(S)=4$ .
"At least one head" excludes only $TT$ : $A = \{HH,\ HT,\ TH\}$ .
Count: $n(A)=3$ .

Answer:

S=\{HH,HT,TH,TT\}

n(S)=4

;

A=\{HH,HT,TH\}

n(A)=3

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Sample space when a single die is rolled?
2.
How many outcomes when two dice are rolled?
3.
A coin is tossed 3 times. How many outcomes?
4.
On one die, list the event "an even number".

An event is a SET of outcomes, not a single outcome

"Getting a number greater than 4" on a die is the event

\{5,6\}

— two outcomes — not one. Mis-reading an event as a single outcome is the most common source of a wrong favourable-count.

Write or size the sample space before you count anything

Almost every classical-probability error is a counting error in

n(S)

n(E)

. Fix

n(S)

first (it is

2^n

for coins,

6^k

for dice,

\binom{n}{r}

for unordered selections), then count favourable outcomes against it.

Drill 1 more on what is probability? (random experiments, sample space, events)

Concept 2 of 8

Classical (theoretical) probability

Intuition

When every outcome is equally likely, probability is just a counting ratio: how many outcomes make the event happen, out of how many are possible. No experiment needed — pure counting.

Definition

For a finite sample space $S$ of equally likely outcomes, the probability of an event $E$ is $P(E) = \dfrac{n(E)}{n(S)}$ — favourable outcomes over total outcomes. Because $0 \le n(E) \le n(S)$ , every probability lies between $0$ and $1$ . This is the *classical* (or theoretical) definition; it fails the moment the outcomes are not equally likely (a loaded die), where each outcome must be weighted instead.

Classical probability

P(E) = \dfrac{n(E)}{n(S)} = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}

$n(E)$ number of outcomes that make $E$ occur
$n(S)$ total number of equally likely outcomes

Worked example

A box contains 5 red, 4 white, and 3 black balls. One ball is drawn at random. What is the probability that it is red?

Total outcomes: $n(S) = 5 + 4 + 3 = 12$ (each ball equally likely).
Favourable: there are 5 red balls, so $n(E) = 5$ .
Divide: $P(\text{red}) = \dfrac{5}{12}$ .

Answer:

\dfrac{5}{12}

Practice this conceptself-check · 4 quick reps

Try it yourself

A single fair die is rolled. What is the probability of getting a number greater than 4?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$P(\text{head})$ for a fair coin?
2.
$P(\text{a number} > 4)$ on a die?
3.
$P(\text{an ace})$ from 52 cards?
4.
$P(\text{a red card})$ from 52 cards?

From the bank · past-year question

Example 2ProbabilityEASY

A card is drawn from a well-shuffled ordinary deck of 52 cards. What is the probability that it is an ace?

[Q111 · Apr · 2017]

The classical formula needs EQUALLY LIKELY outcomes

P=n(E)/n(S)

is only valid when every outcome is equally likely. A loaded die, a biased coin, or faces repeated unevenly (a die with two 4s) breaks the assumption — weight each outcome by its own probability instead of counting.

Favourable is a subset of total, so $P$ can never exceed 1

If a computation gives

P>1

P<0

, the favourable or total count is wrong. Recount before trusting the answer.

Drill 6 more on classical (theoretical) probability

Concept 3 of 8

Axioms, range, complement, and odds

Intuition

Probabilities live between 0 and 1; the whole sample space has probability 1; and the chance an event does NOT happen is 1 minus the chance it does. That last fact — the complement rule — is the single most useful shortcut in the chapter: when an event is messy to count directly ("at least one…"), count its opposite instead.

Definition

Range + axioms: for any event $E$ , $0 \le P(E) \le 1$ , with $P(S)=1$ (certain) and $P(\varnothing)=0$ (impossible).
Complement: $E'$ (" $E$ does not occur") satisfies $P(E') = 1 - P(E)$ .
Odds: odds in favour of $E$ of $a:b$ give $P(E) = \dfrac{a}{a+b}$ ; odds against of $a:b$ give $P(E) = \dfrac{b}{a+b}$ .

Complement rule and range

0 \le P(E) \le 1, \qquad P(S)=1, \qquad P(E') = 1 - P(E)

$E'$ complement of $E$ — the event that $E$ does not occur
$P(\varnothing)$ probability of the impossible event, equal to $0$

Worked example

The probability that it rains tomorrow is

0.3

. (i) What is the probability it does not rain? (ii) If the odds in favour of an event are

3:2

, what is its probability?

(i) Complement: $P(\text{no rain}) = 1 - 0.3 = 0.7$ .
(ii) Odds in favour $3:2$ mean 3 favourable parts out of $3+2=5$ : $P = \dfrac{3}{5}$ .

Answer:(i)

0.7

; (ii)

\dfrac{3}{5}

Practice this conceptself-check · 4 quick reps

Try it yourself

When two dice are thrown, the probability of getting a sum of 7 is

\dfrac{1}{6}

. What is the probability of NOT getting a sum of 7?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$P(E)=0.4$ . Find $P(E')$ .
2.
Odds in favour $2:3$ . Find $P(E)$ .
3.
Odds against an event $5:1$ . Find $P(E)$ .
4.
Can $P(E)=1.2$ ?

From the bank · past-year question

Example 3ProbabilityEASY

In a cricket match, a batsman hits a six 8 times out of 60 balls he plays. What is the probability that on a ball played he does not hit a six?

[Q105 · Sep · 2021]

Odds are not probability: $a:b$ in favour means $\dfrac{a}{a+b}$ , not $\dfrac{a}{b}$

Convert odds to a probability by putting the favourable parts over the TOTAL parts. Odds in favour

3:2

\tfrac{3}{5}

; odds against

3:2

\tfrac{2}{5}

"At least one…" almost always means use the complement

P(\text{at least one}) = 1 - P(\text{none})

. Counting the cases directly (exactly one, exactly two, …) is slower and a frequent source of arithmetic slips.

Drill 2 more on axioms, range, complement, and odds

Concept 4 of 8

Selection probability with combinations

Intuition

When you draw a handful of objects "at random" and order does not matter, both the favourable and the total counts are combinations

\binom{n}{r}

. The probability is one combination count divided by another.

Definition

Choosing $r$ objects from $n$ where order is irrelevant has $\binom{n}{r}$ total outcomes. If the objects split into types (say $a$ of one kind and $b$ of another) and you want $k$ of the first kind, the favourable count is $\binom{a}{k}\binom{b}{r-k}$ . Then $P = \dfrac{\binom{a}{k}\binom{b}{r-k}}{\binom{a+b}{r}}$ .

Selection probability (combinations)

P = \dfrac{\dbinom{a}{k}\dbinom{b}{r-k}}{\dbinom{a+b}{r}}

$a, b$ counts of the two types of object
$k$ how many of the first type the event requires
$r$ total number drawn

Worked example

A bag has 5 red and 4 blue balls. Three balls are drawn at random. What is the probability that exactly 2 are red?

Total ways to draw 3 from 9: $\binom{9}{3} = 84$ .
Favourable: choose 2 red from 5 and 1 blue from 4: $\binom{5}{2}\binom{4}{1} = 10 \times 4 = 40$ .
Divide: $P = \dfrac{40}{84} = \dfrac{10}{21}$ .

Answer:

\dfrac{10}{21}

Practice this conceptself-check · 4 quick reps

Try it yourself

An urn contains 4 white and 6 black balls. Two balls are drawn at random. What is the probability that both are black?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\binom{5}{2} = ?$
2.
Choose 2 from 6 — total ways?
3.
Bag of 3 red, 2 green; draw 2. $P(\text{both red})$ ?
4.
From 5 people, a 2-person committee is chosen. $P(\text{2 specified people})$ ?

From the bank · past-year question

Example 4ProbabilityEASY

An urn contains 10 white and 5 red balls. If two balls are drawn at random, then what is the probability that both the balls are red?

[Q104 · Sep · 2025]

"Drawn together / selected at random" means order does NOT matter — use $\binom{n}{r}$ , not permutations

If you accidentally use

{}^{n}P_r

(ordered) in both numerator and denominator the ratio often still works, but mixing one ordered count with one unordered count gives the wrong answer. Keep both counts unordered.

"At least one of a type" is fastest via the complement

P(\text{at least one red}) = 1 - P(\text{no red}) = 1 - \dfrac{\binom{b}{r}}{\binom{a+b}{r}}

. Summing exactly-one, exactly-two, … wastes time.

Drill 16 more on selection probability with combinations

Concept 5 of 8

Probability with dice

Intuition

Two dice produce

36

equally-likely ordered outcomes

(1,1)

through

(6,6)

. Most dice questions are just "count the

(a,b)

pairs that satisfy the condition, divide by 36". A

6 \times 6

grid of pairs is the fastest way to count sums and differences.

Definition

One die: $S=\{1,\dots,6\}$ , $n(S)=6$ .
Two dice: $n(S)=6^2=36$ ordered pairs; the sum ranges $2$ to $12$ with $7$ the most likely (6 ways).
** $k$ dice:** $n(S)=6^k$ .
Non-standard or loaded die (faces repeated, or weighted) no longer has equally likely faces — weight each face by its own probability rather than using $n(E)/6$ .

Two-dice sample space

n(S) = 6^2 = 36, \qquad P(\text{sum}=7) = \dfrac{6}{36} = \dfrac{1}{6}

$(a,b)$ ordered pair: $a$ on the first die, $b$ on the second

Visualization · two-dice sample space

Target sum: 7

favourable = 6P(sum = 7) = 6/36 = 1/6

Each of the 36 cells is one equally-likely ordered outcome (first die, second die). The highlighted anti-diagonal is the event "sum = 7"; its size over 36 is the probability. The count peaks at 6 for a sum of 7 and tapers to 1 at sums 2 and 12.

Worked example

Two fair dice are thrown. What is the probability that the sum of the numbers is 9 or more?

Total outcomes: $36$ .
Count pairs by sum: sum $9$ : 4 ways; sum $10$ : 3; sum $11$ : 2; sum $12$ : 1. Total $= 10$ .
Divide: $P = \dfrac{10}{36} = \dfrac{5}{18}$ .

Answer:

\dfrac{5}{18}

Practice this conceptself-check · 4 quick reps

Try it yourself

Two fair dice are thrown. What is the probability that the sum is a multiple of 4?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$n(S)$ for two dice?
2.
$P(\text{sum}=7)$ with two dice?
3.
$P(\text{a doublet})$ with two dice?
4.
$P(\text{sum}=12)$ with two dice?

From the bank · past-year question

Example 5ProbabilityEASY

Two fair dice rolled. P(sum=7)?

[Q114 · Apr · 2018]

Two-dice outcomes are ORDERED pairs: $(2,3)$ and $(3,2)$ are different

The sample space has

36

outcomes, not

21

. Counting unordered pairs undercounts every non-doublet event by a factor of two.

Loaded or non-standard dice: faces are NOT equally likely

If a die has two faces showing 4, or even faces are twice as likely as odd, you cannot use

n(E)/36

. Assign each face its probability first, then add up the favourable outcomes' probabilities.

Drill 19 more on probability with dice

Concept 6 of 8

Probability with coins

Intuition

Tossing a fair coin

n

times gives

2^n

equally-likely ordered sequences. For a biased coin with

P(\text{head})=p

, multiply

p

and

1-p

along the sequence. "At least one head" is almost always fastest through the complement.

Definition

**Fair coin, $n$ tosses:** $n(S)=2^n$ , every sequence equally likely; exactly $k$ heads occurs in $\binom{n}{k}$ of them.
**Biased coin, $P(H)=p$ :** a specific sequence with $h$ heads and $t$ tails has probability $p^{h}(1-p)^{t}$ .
At least one head in $n$ fair tosses: $1-\left(\tfrac{1}{2}\right)^{n}$ .

Coin tosses

n(S)=2^{n}, \qquad P(\text{at least one head in } n) = 1 - \left(\tfrac{1}{2}\right)^{n}

$n$ number of tosses
$p$ probability of a head on one toss ( $\tfrac{1}{2}$ if fair)

Diagram · two coin tosses → 2² = 4 outcomes

Each toss branches into H or T with probability ½, and the branches multiply: every leaf is ½ × ½ = ¼. Tossing n coins gives 2ⁿ equally likely outcomes — so "at least one head" is easiest via the complement, 1 − P(all tails).

Worked example

A fair coin is tossed 3 times. What is the probability of getting at least one head?

Total sequences: $2^3 = 8$ .
Complement "no head" is the single sequence $TTT$ : $P(\text{no head}) = \dfrac{1}{8}$ .
So $P(\text{at least one head}) = 1 - \dfrac{1}{8} = \dfrac{7}{8}$ .

Answer:

\dfrac{7}{8}

Practice this conceptself-check · 4 quick reps

Try it yourself

A fair coin is tossed 4 times. What is the probability of getting exactly two heads?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$n(S)$ for 3 coin tosses?
2.
Fair coin, 3 tosses: $P(\text{all heads})$ ?
3.
Fair coin, 2 tosses: $P(\text{at least one tail})$ ?
4.
Biased coin $P(H)=\tfrac{1}{3}$ : $P(HH)$ in 2 tosses?

From the bank · past-year question

Example 6ProbabilityEASY

A biased coin with the probability of getting head equal to

\frac{1}{4}

is tossed five times. What is the probability of getting tail in all the first four tosses followed by head?

[Q103 · Apr · 2023]

Sequences are ordered: $HT \ne TH$

Count the

2^n

ordered sequences, not the

n+1

"number of heads" buckets — those buckets are not equally likely ("exactly 1 head in 2 tosses" has 2 sequences, "2 heads" has 1).

Biased coin: do not use $2^n$ equally-likely counting

When

P(H) \ne \tfrac{1}{2}

, each sequence has its own probability

p^{h}(1-p)^{t}

. Counting outcomes and dividing by

2^n

is only valid for a fair coin.

Drill 6 more on probability with coins

Concept 7 of 8

Probability with arrangements

Intuition

When people or letters are arranged in a row "at random", the total is

n!

arrangements (divided by factorials for repeats). For "

X

and

Y

together", glue them into a single block. Probability is favourable arrangements over total arrangements.

Definition

Total arrangements of $n$ distinct objects: $n!$ . For two specified objects to be together, treat them as one block: that block plus the other $n-2$ objects is $(n-1)!$ arrangements, times $2!$ for the block's internal order. So $P(\text{two specified together}) = \dfrac{2\,(n-1)!}{n!} = \dfrac{2}{n}$ . With repeated letters, divide $n!$ by the factorial of each repeat count.

Arrangement probability (two together)

P(\text{two specified together}) = \dfrac{2\,(n-1)!}{n!} = \dfrac{2}{n}

$n$ number of objects being arranged
$2\,(n-1)!$ favourable: glue the pair ( $(n-1)!$ ) and order them internally ( $2!$ )

Worked example

Five people sit in a row at random. What is the probability that two particular people, A and B, sit together?

Total arrangements: $5! = 120$ .
Glue A and B into one block: the block plus 3 others give $4!$ arrangements, and the block can be $AB$ or $BA$ : favourable $= 2 \times 4! = 48$ .
Divide: $P = \dfrac{48}{120} = \dfrac{2}{5}$ . (Matches $\tfrac{2}{n}=\tfrac{2}{5}$ .)

Answer:

\dfrac{2}{5}

Practice this conceptself-check · 4 quick reps

Try it yourself

The letters of the word DELHI are arranged at random. What is the probability that the two vowels E and I are together?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Total arrangements of 4 distinct people?
2.
4 people in a row: $P(\text{A, B together})$ ?
3.
Number of arrangements of the letters in BOOK?
4.
6 people in a row: $P(\text{A, B together})$ ?

From the bank · past-year question

Example 7ProbabilityEASY

In a class of

n

students including P and Q, what is the probability that P and Q sit together if seats assigned randomly?

[Q114 · Sep · 2023]

"Together" = glue into a block, then multiply by the block's internal arrangements

Forgetting the

2!

for the order within the pair halves the favourable count. For a block of

k

people the internal factor is

k!

Repeated letters divide the total by the factorial of each repeat count

The arrangements of TIRUPATI use

\dfrac{8!}{2!\,2!}

(two T's, two I's), not

8!

. Forgetting this inflates both counts unequally and breaks the ratio.

Drill 13 more on probability with arrangements

Concept 8 of 8

Choosing numbers with a property

Intuition

Many PYQs choose a number (or a few numbers) at random from a finite set and ask the probability it has some property — consecutive, a multiple, satisfies an inequality. The recipe is unchanged: count the elements with the property, divide by the size of the set (or use

\binom{n}{r}

when several are chosen at once).

Definition

A number chosen at random from $\{1,\dots,n\}$ : $P = \dfrac{\#\{x \text{ with the property}\}}{n}$ . When several numbers are chosen together, the denominator becomes $\binom{n}{r}$ . Useful counts: multiples of $d$ in $\{1,\dots,n\}$ number $\lfloor n/d\rfloor$ ; consecutive triples among $1,\dots,N$ number $N-2$ .

Counting favourable numbers

P = \dfrac{\#\{x : x \text{ has the property}\}}{n}

$\lfloor n/d\rfloor$ how many of $1,\dots,n$ are multiples of $d$

Worked example

A number is chosen at random from 1 to 20. What is the probability that it is a multiple of 3 or a multiple of 5?

Multiples of 3 up to 20: $\lfloor 20/3\rfloor = 6$ . Multiples of 5: $\lfloor 20/5\rfloor = 4$ .
Multiples of both (i.e. of 15): $\lfloor 20/15\rfloor = 1$ . By inclusion-exclusion, favourable $= 6 + 4 - 1 = 9$ .
Divide: $P = \dfrac{9}{20}$ .

Answer:

\dfrac{9}{20}

Practice this conceptself-check · 4 quick reps

Try it yourself

A number is chosen at random from the first 30 natural numbers. What is the probability that it is divisible by 4?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
From $1$ – $10$ , $P(\text{prime})$ ?
2.
From $1$ – $20$ , $P(\text{multiple of 5})$ ?
3.
Consecutive triples among $1,\dots,10$ — how many?
4.
From $1$ – $50$ , $P(\text{divisible by 10})$ ?

From the bank · past-year question

Example 8ProbabilityEASY

Three distinct natural numbers are chosen at random from 1 to 10. What is the probability that they are consecutive?

[Q103 · Sep · 2024]

"Or" on number properties needs inclusion-exclusion

P(\text{mult of 3 or 5}) = P(3) + P(5) - P(15)

. Forgetting to subtract the overlap (multiples of both) double-counts and overstates the probability.

Several numbers chosen at once: denominator is $\binom{n}{r}$ , not $n$

"Three numbers chosen from 1 to 10" has

\binom{10}{3}=120

outcomes. Only use

n

in the denominator when exactly one number is picked.

Drill 16 more on choosing numbers with a property

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (7)

Classical (theoretical) probability
Classical probability
$P(E) = \dfrac{n(E)}{n(S)} = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$
Axioms, range, complement, and odds
Complement rule and range
$0 \le P(E) \le 1, \qquad P(S)=1, \qquad P(E') = 1 - P(E)$
Selection probability with combinations
Selection probability (combinations)
$P = \dfrac{\dbinom{a}{k}\dbinom{b}{r-k}}{\dbinom{a+b}{r}}$
Probability with dice
Two-dice sample space
$n(S) = 6^2 = 36, \qquad P(\text{sum}=7) = \dfrac{6}{36} = \dfrac{1}{6}$
Probability with coins
Coin tosses
$n(S)=2^{n}, \qquad P(\text{at least one head in } n) = 1 - \left(\tfrac{1}{2}\right)^{n}$
Probability with arrangements
Arrangement probability (two together)
$P(\text{two specified together}) = \dfrac{2\,(n-1)!}{n!} = \dfrac{2}{n}$
Choosing numbers with a property
Counting favourable numbers
$P = \dfrac{\#\{x : x \text{ has the property}\}}{n}$

Watch out for (16)

An event is a SET of outcomes, not a single outcome
→ What is probability? (Random experiments, sample space, events)
Write or size the sample space before you count anything
→ What is probability? (Random experiments, sample space, events)
The classical formula needs EQUALLY LIKELY outcomes
→ Classical (theoretical) probability
Favourable is a subset of total, so $P$ can never exceed 1
→ Classical (theoretical) probability
Odds are not probability: $a:b$ in favour means $\dfrac{a}{a+b}$ , not $\dfrac{a}{b}$
→ Axioms, range, complement, and odds
"At least one…" almost always means use the complement
→ Axioms, range, complement, and odds
"Drawn together / selected at random" means order does NOT matter — use $\binom{n}{r}$ , not permutations
→ Selection probability with combinations
"At least one of a type" is fastest via the complement
→ Selection probability with combinations
Two-dice outcomes are ORDERED pairs: $(2,3)$ and $(3,2)$ are different
→ Probability with dice
Loaded or non-standard dice: faces are NOT equally likely
→ Probability with dice
Sequences are ordered: $HT \ne TH$
→ Probability with coins
Biased coin: do not use $2^n$ equally-likely counting
→ Probability with coins
"Together" = glue into a block, then multiply by the block's internal arrangements
→ Probability with arrangements
Repeated letters divide the total by the factorial of each repeat count
→ Probability with arrangements
"Or" on number properties needs inclusion-exclusion
→ Choosing numbers with a property
Several numbers chosen at once: denominator is $\binom{n}{r}$ , not $n$
→ Choosing numbers with a property

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1ProbabilityEASY

If a coin is tossed till the first head appears, then what will be the sample space ?

[Q109 · Sep · 2019]

Example 2ProbabilityEASY

What is the probability that February of a leap year selected at random, will have five Sundays?

[Q111 · Apr · 2020]

Example 3ProbabilityMODERATE

A card is drawn from a pack of 52 cards. A gambler bets that it is either a spade or an ace. The odds against his winning are

[Q113 · Sep · 2022]

Example 4ProbabilityMODERATE

Five sticks of length 1, 3, 5, 7 and 9 feet are given. Three of these sticks are selected at random. What is the probability that the selected sticks can form a triangle?

[Q119 · Sep · 2017]

Example 5ProbabilityMODERATE

Three fair dice are thrown. What is the probability of getting a total greater than or equal to 15?

[Q109 · Sep · 2022]

Drill every past-year question on this subtopic

85 questions from the bank — paginated, with cart and Word-export support.

Drill the 85 questions

Drill every past-year question on this subtopic

Related notes